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Chemistry Question on Equilibrium

The values of KspK_{sp} of CaCO3CaCO_3 and CaC2O4CaC_2O_4 are 4.7×10−94.7 \times 10^{-9} and 1.3×10−91.3 \times 10^{-9} respectively at 25�C25�C. If the mixture of these two is washed with water, what is the concentration of Ca2+Ca^{2+} ions in water?

A

5.831×10−5 M5.831 \times 10^{-5}\, M

B

6.856×10−5 M6.856 \times 10^{-5}\, M

C

3.606×10−5 M3.606 \times 10^{-5}\, M

D

7.746×10−5 M7.746 \times 10^{-5}\, M

Answer

7.746×10−5 M7.746 \times 10^{-5}\, M

Explanation

Solution

Ksp(CaCO3)=4.7×10−9K _{ sp }\left( CaCO _{3}\right)=4.7 \times 10^{-9} Ksp(CaC2O4)=1.3×10−9K _{ sp }\left( CaC _{2} O _{4}\right)=1.3 \times 10^{-9} On dividing (1) &(2), (S+S1)×S(S+S1)×S1=4.7×10−91.3×10−9\frac{\left(S+S_{1}\right) \times S}{\left(S+S_{1}\right) \times S_{1}}=\frac{4.7 \times 10^{-9}}{1.3 \times 10^{-9}} SS1=4713=3.61\frac{S}{S_{1}}=\frac{47}{13}=3.61 S=3.61S1...(3)S =3.61 S _{1}...(3) From (1), (S+S1)×S=4.7×10−9\left( S + S _{1}\right) \times S =4.7 \times 10^{-9} From (2), (S+S1)×S1=1.3×10−9\left( S + S _{1}\right) \times S _{1}=1.3 \times 10^{-9} (3.61S1+S1)×S1=1.3×10−9\left(3.61 S _{1}+ S _{1}\right) \times S _{1}=1.3 \times 10^{-9} 4.61S12=1.3×10−94.61 S _{1}{ }^{2}=1.3 \times 10^{-9} S12=0.281×10−9S _{1}{ }^{2}=0.281 \times 10^{-9} S1=0.281×10−9=1.67×10−5S _{1}=\sqrt{0.281 \times 10^{-9}}=1.67 \times 10^{-5} From (3), S=3.61S1=3.61×1.67×10−5=6.02×10−5S =3.61 S _{1}=3.61 \times 1.67 \times 10^{-5}=6.02 \times 10^{-5} [Ca2+]=S+S1=(6.02+1.67)×10−5=7.69×10−5\left[ Ca ^{2+}\right]= S + S _{1}=(6.02+1.67) \times 10^{-5}=7.69 \times 10^{-5}